Question

one_count = []
two_count = []
three_count = []
four_count = []
five_count = []
six_count = []
seven_count = []

counter = Counter(hand)
for d in counter:
    if counter[d] == 1:
        one_count.append(d)
    if counter[d] == 2:
        two_count.append(d)
    if counter[d] == 3:
        three_count.append(d)
    if counter[d] == 4:
        four_count.append(d)
    if counter[d] == 5:
        five_count.append(d)
    if counter[d] == 6:
        six_count.append(d)
    if counter[d] == 7:
        seven_count.append(d)

one_dict = Counter(one_count)
two_dict = Counter(two_count)
three_dict = Counter(three_count)
four_dict = Counter(four_count)
five_dict = Counter(five_count)
six_dict = Counter(six_count)
seven_dict = Counter(seven_count)

print one_dict
print two_dict
print three_dict
print four_dict
print five_dict
print six_dict
print seven_dict

你好！运行上面的程序后，其打印的内容是Counter（）七次。我似乎无法弄清楚出了什么问题，也无法弄清楚眼睛是否新鲜，也许有人能够找出出什么问题了。请记住，此代码只是程序的一部分。

编辑：我意识到我可以只使用Counter（hand），而不用嵌套很多if和for语句。谢谢大家的帮助！

Answer 1

无论hand是什么，one_used都是一个空列表，因此您的行

for e in one_used:

对一个空列表进行迭代，换句话说，什么都不做。

Answer 2

您正在定义Private pNumber As Integer Private pProperties As New Collection Private pWealth As Long Property Get Number() As Integer Number = pNumber End Property Property Let Number(pNum As Integer) pNumber = pNum End Property Property Get Properties() As Collection Properties = pProperties End Property Property Get Wealth() As Long Number = pWealth End Property Property Let Wealth(w As Long) pWealth = w End Property Public Sub addProperty(property As PropertyUnit) pProperties.Add (property) End Sub，这是一个空列表。然后，您遍历空白列表，因此one_used = []甚至不存在。这样会抛出所有循环并进行检查，因此没有要追加的内容。

问题：for d in hand: for e in one_used:为空。

为什么.append（）在此程序中不起作用？

2 个答案: